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If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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If \(a^2b^2c^3\) = 4500. Is b+c = 7 ? (1) a, b and c are positive integers (2) a > b what would be the most efficient way to approach this within 2 mins?
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Originally posted by GMATBaumgartner on 17 Oct 2012, 01:52.
Last edited by abhimahna on 27 Apr 2017, 00:12, edited 2 times in total.
Edited the question.



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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17 Oct 2012, 03:56
GMATBaumgartner wrote: If a^2b^2c^3 = 4500. Is b+c = 7 ?
(1) a, b and c are positive integers (2) a > b
what would be the most efficient way to approach this within 2 mins? Given: \(a^2*b^2*c^3 = 4500=2^2*3^2*5^3\). Question: is \(b+c=7\) (1) a, b and c are positive integers > sure it's possible that \(a=3\), \(b=2\) and \(c=5\) and in this case the answer will be YES but it's also possible that \(a=2\), \(b=3\) and \(c=5\) and in this case the answer will be NO. Not sufficient. (2) a > b. Clearly insufficient. (1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient. Answer: E. Hope it's clear.
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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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17 Oct 2012, 04:13
Thank you bunuel. I completely overlooked the part they there is NO mention of then as primes!
Is there any post where you sum up the important such 'overlookables' that we must crosscheck before marking a statement Sufficient or Not sufficient.
Thanks in advance.I am in the final leg of my prep,any advise would be much appreciated.



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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17 Oct 2012, 05:03
Very often I overlook the possibilty of 0 in equalties. i.e. I check for only +ve and ve numbers. A list of the aforementioned "overlookables" would help a lot.
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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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17 Oct 2012, 19:49



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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23 Oct 2012, 06:40



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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19 Jul 2013, 01:22



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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Updated on: 19 Jul 2013, 12:49
Hi,
Sorry I think that answer is B. Here is my opinion: we are just able to rewrite 4500=(3^2)*(2^2)*(5^3), right?
just we don't know that which one of 3 or 2 is a & b, correct?
statement 2 is giving us a>b then we know that a=3 and b=2 and problem solved.
I don't know where I went wrong? Can you explain my fault?
Edit: I found it!
Originally posted by zazoz on 19 Jul 2013, 12:41.
Last edited by zazoz on 19 Jul 2013, 12:49, edited 1 time in total.



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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19 Jul 2013, 12:47
zazoz wrote: Hi,
Sorry I think that answer is B. Here is my opinion: we are just able to rewrite 4500=(3^2)*(2^2)*(5^3), right?
just we don't know that which one of 3 or 2 is a & b, correct?
statement 2 is giving us a>b then we know that a=3 and b=2 and problem solved.
I don't know where I went wrong? Can you explain my fault? Please read the solution: ifa2b2c34500isbc140842.html#p1132262Pay attention to the examples given: (1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient. Answer: E. Hope it's clear.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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19 Jul 2013, 12:49
Bunuel wrote: zazoz wrote: Hi,
Sorry I think that answer is B. Here is my opinion: we are just able to rewrite 4500=(3^2)*(2^2)*(5^3), right?
just we don't know that which one of 3 or 2 is a & b, correct?
statement 2 is giving us a>b then we know that a=3 and b=2 and problem solved.
I don't know where I went wrong? Can you explain my fault? Please read the solution: ifa2b2c34500isbc140842.html#p1132262Pay attention to the examples given: (1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient. Answer: E. Hope it's clear. yeah, I got it, thanks



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Re: If a^2b^2c^3 = 4500. Is b+c = 7 ? [#permalink]
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16 Mar 2016, 21:56
If a^2b^2c^3 = 4500. Is b+c = 7 ? (a^2) (b^2) (c^3) =4500=(30^2) 5 (abc)^2 c= (30^2) 5 If a, b and c are positive integers then we may have c=5 and abc=30 => ab=6
(1) a, b and c are positive integers a, b and c are positive integers then we may have c=5 and abc=30 => ab=6. but a=3,b=2 => b+c=7 a=6, b=1 => b+c=6 INSUFFICIENT (2) a > b a and b may be fractions and many combinations are posssible for a and b. INSUFFICIENT
combining 1 and 2 a=3,b=2 => b+c=7 a=6, b=1 => b+c=6 INSUFFICIENT
Hence E Thanks



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